Complex roots of equation with geometric sequence?

Question

1 + z + z^2 + z^3 = 0

How to find the solutions to this one?

My workbook tells me to use the fact that it's a geometric sequence, but I haven't worked with them at all, so I am not sure I know what to do with that hint.

Answer 1

Sum of this is: $$1+x+x^2+x^3+\cdots+x^n=\frac{x^{n+1}-1}{x-1}$$ Try expanding: $$(x-1)(1+x+x^2+\cdots+x^n)$$ Or use long division: $$\begin{array}{|c|c|c|c} x-1&x^{n+1}-1&x^n\\ &x^{n+1}-x^n\\\hline x-1&x^n-1&x^{n-1}\\ &x^n-x^{n-1}\\\hline \cdots&\cdots&\cdots\\ \cdots&\cdots&\cdots\\\hline x-1&x-1&1\\ &x-1\\\hline &0 &\end{array}$$ So: $$1+x+x^2+x^3=\frac{x^4-1}{x-1}\quad x\ne1$$ So: $$x^4=1\implies x=\pm i,-1$$

Answer 2

$$\begin{align}1+z+z^2+z^3&=0\\ (1+z+z^2+z^3+\color{red}{z^4+z^5+z^6+\cdots} )-(\color{red}{z^4+z^5+z^6+\cdots})&=0\\ \color{blue}{(1+z+z^2+z^3+\cdots )}-z^4\color{blue}{(1+z+z^2+z^3+\cdots )}&=0\\ (1-z^4)\color{blue}{(1+z+z^2+z^3+\cdots )}&=0\\ (1-z^4)\cdot \color{blue}{\frac 1{1-z}}&=0\\ z^4&=1 \quad (z\neq1 )\\ z&=i,-1,-i \end{align}$$

Or, more correctly, $$z=\left(2n+\frac r2\right)\pi$$ where $n=0,\pm1, \pm2, \cdots $ and $r=1,2,3$.